Find the energy and mometum of a photon of ultraviolet radiation of 280 nm wavelength.

Explanation

To find the energy and momentum of a photon of ultraviolet radiation with a wavelength of 280 nm, we can follow these steps:

### Step 1: Convert the wavelength from nanometers to meters
The given wavelength is 280 nm. To convert this to meters, we use the conversion factor \(1 \text{ nm} = 10^{-9} \text{ m}\).

\[ \lambda = 280 \text{ nm} = 280 \times 10^{-9} \text{ m} = 2.80 \times 10^{-7} \text{ m} \]

### Step 2: Calculate the energy of the photon
The energy \(E\) of a photon can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \(h\) (Planck's constant) = \(6.626 \times 10^{-34} \text{ J s}\)
- \(c\) (speed of light) = \(3.0 \times 10^{8} \text{ m/s}\)
- \(\lambda\) = \(2.80 \times 10^{-7} \text{ m}\)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34} \text{ J s})(3.0 \times 10^{8} \text{ m/s})}{2.80 \times 10^{-7} \text{ m}} \]

Calculating this gives:

\[ E = \frac{1.9878 \times 10^{-25} \text{ J m}}{2.80 \times 10^{-7} \text{ m}} \approx 7.09 \times 10^{-19} \text{ J} \]

To convert this energy from joules to electron volts (1 eV = \(1.6 \times 10^{-19} \text{ J}\)):

\[ E \approx \frac{7.09 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 4.43 \text{ eV} \]

### Step 3: Calculate the momentum of the photon
The momentum \(p\) of a photon can be calculated using the formula:

\[ p = \frac{h}{\lambda} \]

Substituting the values:

\[ p = \frac{6.626 \times 10^{-34} \text{ J s}}{2.80 \times 10^{-7} \text{ m}} \]

Calculating this gives:

\[ p \approx 2.37 \times 10^{-27} \text{ kg m/s} \]

### Final Results
- The energy of the photon is approximately \(4.43 \text{ eV}\).
- The momentum of the photon is approximately \(2.37 \times 10^{-27} \text{ kg m/s}\).

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